Prove the following set is uncountable

Question

I need some help with part (b) of the question. Would appreciate feedback on whether my solution to part (a) is correct too.

For each ∈ ℝ, define = { + ∶ ∈ ℤ}. Let = { ∶ ∈ ℝ}.

(a) Prove that is countable for every ∈ ℝ.
(b) Prove that is uncountable.

You may use without proof the fact that a set is countable if and only if there is a sequence 0, 1, 2, … ∈ in which every element of appears.

For part (a), I proved that for all ∈ ℝ, can be written in a sequence defined as below:

c2i = x - i
c2i+1 = x + i + 1

i.e. = {x, x + 1, x - 1, x + 2, x - 2, x + 3, x - 3, ....}

For part (b), however, I am stuck on proving . I believe it is probably something to do with cardinality of Unions? Since is just a Union of 1, 2, 3 ...

However, in my current syllabus, one theorem I am taught is that:
"Let A,B be countable infinite sets. Then A U B is countable."

Thank you for taking the time to read this and I appreciate all feedback! Thank you

Answer 1

For the solution on part a, the idea is correct, but there is a typo in $c_{2i}=x-i$.

For part (b): $[0,1)$ can be imbedded in $\mathscr{C}$ via the map $x\mapsto Ax$.

Let $x,y\in[0,1)$ with $Ax=Ay$, then there must be some $k\in\mathbb{Z}$ with $x=y+k$. As $|x-y|<1$, thus $k=0$ and $x=y$.

This proves the injectivity. $[0,1)$ is uncountable and so is $\mathscr{C}$.

Answer 2

For $b)$ we have that $Ax$ is distinct between $A(0)$ and $A(1)$. Hence the cardinality is at least the same as the interval $(0,1)$, which we know is uncountable.

Answer 3

Since is just a Union of 1, 2, 3 ...

No! is a set of sets, not a union of sets. The cardinality of is just the number of sets it contains. Naively you might think that this is the same as the cardinality of $\{x:x\in\Bbb R\}$, which of course is just $\mathfrak{c}$. But there are duplications in there, so you have to be a bit careful.